5 Steps to a 5 AP Chemistry

17.An electrolysis cell was constructed with two
platinum electrodes in a 1.00 M aqueous solu-
tion of KCl. An odorless gas evolves from one
electrode, and a gas with a distinctive odor
evolves from the other electrode. Choose the cor-
rect statement from the following list.

(A) The gas with the distinctive odor was evolved

at the anode.

(B) The odorless gas was oxygen.

(C) The gas with the distinctive odor was evolved

at the negative electrode.

(D) The odorless gas was evolved at the positive

electrode.

(E) The odorless gas was evolved at the anode.

18.H 2 O 2 (aq) +KIO 4 (aq) →KIO 3 (aq) +O 2 (g) +
H 2 O(l)

Choose the true statement from the following list.

(A) The iodine oxidation state is reduced from

+8 to +6.

(B) This is not an oxidation–reduction reaction.

(C) H 2 O 2 behaves as a reducing agent.

(D) Hydrogen is reduced from +2 to +1.

(E) H 2 O 2 behaves as an oxidizing agent.

Questions 19 and 20 are concerned with the fol-
lowing half-reaction in an electrolytic cell:

2 BrO 3 −+12 H++10 e−→Br 2 +6 H 2 O

19.Choose the correct statement from the following

list.

(A) The BrO 3 −undergoes oxidation at the anode.

(B) Br goes from a −1 oxidation to a 0 oxidation

state.

(C) Br 2 is oxidized at the anode.

(D) H+is a catalyst.

(E) The BrO 3 − undergoes reduction at the

cathode.

20.If a current of 5.0 A is passed through the elec-

trolytic cell for 0.50 h, how should you calculate

the number of grams of Br 2 that will form?

(A) (5.0)(0.50)(3600)(159.8)/(10)

(B) (5.0)(0.50)(3600)(159.8)/(96500)(10)

(C) (5.0)(0.50)(60)(159.8)/(96500)(10)

(D) (5.0)(0.50)(3600)(79.9)/(96500)(10)

(E) (5.0)(0.50)(159.8)/(96500)(10)

21.2 M(s) +3 Zn^2 +(aq) →2 M^3 +(aq) + 3 Zn^2 +(aq)

E°=0.90 V

Zn^2 +(aq) +2e− →Zn(s) E°=−0.76 V

Using the above information, determine the stan-

dard reduction potential for the following reaction:

(A) 0.90 V

(B) +1.66 V

(C) 0.00 V

(D) −0.62 V

(E) −1.66V

Maq e Ms^3 +−()+→ 3 ()

Electrochemistry  253

 Answers and Explanations

1. D—The addition of zinc ion, from the ZnSO 4 ,
   increases the zinc concentration. This increases
   the numerator in the logarithm part of the
   Nernst equation. This is a negative term, so the
   cell voltage will decrease.


2. A—The size of the electrode is not important.


3. B—The salt bridge serves as an ion source to
   maintain charge neutrality. Deionized water
   would not be an ion source, so the cell could not
   operate.
   4. D—As the cell operates, the copper ion concen-
   tration would decrease and the zinc ion concen-
   tration would increase. Both of these changes
   would make the logarithm term in the Nernst
   equation more negative. This would decrease the
   voltage.
   5. A—The balanced equation is:

2165

28

424

2

2

MnO aq H aq C O aq

Mn aq

−+ −

+

++

→+

() () ()

() HHOl 22 ()+ 10 CO g( )